Math Skills Review IMath Skills Review I

Subjects include:Subjects include:•Algebra I/IIAlgebra I/II•GeometryGeometry

•TrigonometryTrigonometry

Christopher BullardChristopher Bullard4/17/20074/17/2007

Question #1Question #1

John is at a dealership to buy a used car. There John is at a dealership to buy a used car. There are two vehicles he is choosing from (an ’86 four are two vehicles he is choosing from (an ’86 four door sedan, and an ’89 two door coupe). All cars door sedan, and an ’89 two door coupe). All cars at the dealership have a 10% mark-up of their fair at the dealership have a 10% mark-up of their fair market value. The ’89 coupe has a dealership market value. The ’89 coupe has a dealership price $500 more than the fair market value of the price $500 more than the fair market value of the ’86 sedan. The ’86 sedan’s dealership price is ’86 sedan. The ’86 sedan’s dealership price is $4,950.$4,950.

What is the fair market value of the ’89 sedan?What is the fair market value of the ’89 sedan?

Solution to Question #1Solution to Question #1

First, we must determine the fair market value of the ’86 sedan. First, we must determine the fair market value of the ’86 sedan. To do this, we set up the equation :To do this, we set up the equation :

1.1x=49501.1x=4950

where x represents the ’86 sedan’s fair market valuewhere x represents the ’86 sedan’s fair market value

Solving for this yields x=$4,500Solving for this yields x=$4,500

Next, we know that the ’89 sedan’s dealership price is $500 more Next, we know that the ’89 sedan’s dealership price is $500 more than the ’86 sedan’s fair market value. Therefore, we set up the than the ’86 sedan’s fair market value. Therefore, we set up the equation:equation:

x+500=1.1yx+500=1.1y

We already know what x is, so solving for y, we find the fair value We already know what x is, so solving for y, we find the fair value of the ’89 sedan to be $4,955, rounded to the nearest dollar.of the ’89 sedan to be $4,955, rounded to the nearest dollar.

Question #2Question #2

A box has a volume of 2,000 cmA box has a volume of 2,000 cm33 (cubic centimeters, where 2.54 (cubic centimeters, where 2.54 cm = 1 inch), and the vertical surface adjacent to A has surface cm = 1 inch), and the vertical surface adjacent to A has surface area of 250 cmarea of 250 cm22 (10cm x 25cm). Given angle A= 30 degrees, find (10cm x 25cm). Given angle A= 30 degrees, find the length of the line AB.the length of the line AB.

Solution to Question #2Solution to Question #2

First, our objective is to solve for the hypotenuse of the right triangle First, our objective is to solve for the hypotenuse of the right triangle formed on the “floor” of the box (the right triangle with adjacent side of formed on the “floor” of the box (the right triangle with adjacent side of length 25cm, and opposite side of length 8cm). We can either use trig length 25cm, and opposite side of length 8cm). We can either use trig functions (sine, or cosine), or the Pythagorean Theorem. Either way, the functions (sine, or cosine), or the Pythagorean Theorem. Either way, the result is the square root of 689 (we will leave it in this form until after we result is the square root of 689 (we will leave it in this form until after we have proceeded with the second step).have proceeded with the second step).

The next step involves solving for the right triangle now formed by this new The next step involves solving for the right triangle now formed by this new side (call it AF), and the height of the box (10cm). Solving for this right side (call it AF), and the height of the box (10cm). Solving for this right triangle will yield AB’s length of 28.089cm (the square root of 789).triangle will yield AB’s length of 28.089cm (the square root of 789).

Note, however, that we could also solve this problem if only given the Note, however, that we could also solve this problem if only given the surface area of the 25cm by 10cm side, and the total volume of the box. surface area of the 25cm by 10cm side, and the total volume of the box. We could obtain the length of 8cm by dividing the volume by the surface We could obtain the length of 8cm by dividing the volume by the surface area, and then obtain the hypotenuse of the “floor” of the box by the area, and then obtain the hypotenuse of the “floor” of the box by the cotangent of 30 degrees, multiplied by division of volume by area.cotangent of 30 degrees, multiplied by division of volume by area.

Question #3Question #3

Find the value of the variable x from the equationFind the value of the variable x from the equation

37 = 3 + 47 log37 = 3 + 47 logxx8080

Question #4Question #4

Find the value of the variable y from the equationFind the value of the variable y from the equation

42 = 100 – 6 log42 = 100 – 6 log77yy

Question #5Question #5

Find the value of the variable z from the equationFind the value of the variable z from the equation

91 = ln (13/z)91 = ln (13/z)

Solution to Question #3Solution to Question #3

Given logGiven logaab = (log b)/(log a)b = (log b)/(log a)

(34/47) = (log(34/47) = (log101080)/ (log80)/ (log1010x)x)so…so…

x = 10x = 10[(log 80)/(34/47)][(log 80)/(34/47)]

Solution to Question #4Solution to Question #4Given logGiven logaab = (log b)/(log a)b = (log b)/(log a)

(-58/6) = (log y)/ (log 7)(-58/6) = (log y)/ (log 7)

so…so…

y = 10y = 10(-58/6) (log7)(-58/6) (log7)

Solution to Question #5Solution to Question #5

Given ln(a/b) = ln(a) – ln (b)Given ln(a/b) = ln(a) – ln (b)

ln (13) - 91 = ln (z)ln (13) - 91 = ln (z)

So…So…

z = e z = e (ln (13) - 91 )(ln (13) - 91 )

Question #6Question #6

Given the following equationsGiven the following equations

(1) 3w – 2x + 7y +z = 108(1) 3w – 2x + 7y +z = 108

(2) w + x – 2y +3z = 372(2) w + x – 2y +3z = 372

(3) 2w + z = 13(3) 2w + z = 13

(4) x + y = 7(4) x + y = 7

Solve for w, x, y, and z, listing them in the form (w, x, y, z).Solve for w, x, y, and z, listing them in the form (w, x, y, z).

Solution to Question #6Solution to Question #6We must perform either row operations or substitution with equations (3) and (4) to simplify what We must perform either row operations or substitution with equations (3) and (4) to simplify what we have. For this example, we will begin with row operations to eliminate unnecessary variables. we have. For this example, we will begin with row operations to eliminate unnecessary variables. For convenience, we will eliminate either w or z from equation (2) or (1), respectively. This results For convenience, we will eliminate either w or z from equation (2) or (1), respectively. This results in…in…

(1’) x + y +10z = 1224(1’) x + y +10z = 1224

(2) w + x – 2y +3z = 372(2) w + x – 2y +3z = 372

(3) 2w + z = 13(3) 2w + z = 13

(4) x + y = 7(4) x + y = 7

We can now use equation (4) to reduce the equation (1’) to…We can now use equation (4) to reduce the equation (1’) to…

(7 – y) + y +10z = 1224(7 – y) + y +10z = 1224

which now lets us determine…which now lets us determine…

7 + 10z = 12247 + 10z = 1224

where z = 121.7where z = 121.7

Next, since we know z, we can substitute into equation (3) to determine…Next, since we know z, we can substitute into equation (3) to determine…

2w = 13 – 121.72w = 13 – 121.7

Which results in…Which results in…

w = -54.35w = -54.35

Finally, given w and z, we can use equations (2) and (4) to determine y, which results in…Finally, given w and z, we can use equations (2) and (4) to determine y, which results in…

(2) -54.35 + x – 2y +3(121.7) = 372, or…(2) -54.35 + x – 2y +3(121.7) = 372, or…

310.75 + x – 2y = 372… where we can substitute equation (4) to get…310.75 + x – 2y = 372… where we can substitute equation (4) to get…(Continued on next slide!)(Continued on next slide!)

Solution to Question #6 (continued)Solution to Question #6 (continued)Finally, given w and z, we can use equations (2) and (4) to determine y, which results in…Finally, given w and z, we can use equations (2) and (4) to determine y, which results in…

(2) -54.35 + x – 2y +3(121.7) = 372, or…(2) -54.35 + x – 2y +3(121.7) = 372, or…

310.75 + x – 2y = 372… where we can substitute equation (4) to get…310.75 + x – 2y = 372… where we can substitute equation (4) to get…

310.75 + (7 – y) -2y = 372, or…310.75 + (7 – y) -2y = 372, or…

-3y = 54.25, leaving y = -18.08333-3y = 54.25, leaving y = -18.08333

and x = 7 – y, or x = 7 + 18.08333and x = 7 – y, or x = 7 + 18.08333

Leaving the solution to the system of equations to be…Leaving the solution to the system of equations to be…

(-54.35, 25.08333, -18.08333, 121.7)(-54.35, 25.08333, -18.08333, 121.7)

End of Math Skills Review IEnd of Math Skills Review IContact information:Contact information:

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